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Short answere here is: Where you cannot use anything else. In C you don't have any support for complex datatypes such as a string. There are also no way of passing a variable "by reference" to a function. That's where you have to use pointers. Also you can have them to point at virtually anything, linked lists, members of structs and so on. Buty let's not go into that here.

The pointer operator (the *) is needed since we are telling printf that we want to print a character. Without the *, the character representation of the memory address itself would be printed. Now we are using the character itself instead. If we had used %s instead of %c, we would have asked prinf to print the content of the memory address pointed to by 'a' plus one (in this example above), and we wouldn't have had to put the * in front:

But this would not have just printed the second character, but instead all characters in the nextcoming memory addresses, until a null character (\0) were found. And this is where things start to get dangerous. What if you accidentally try and print a variable of the type integer instead of a char pointer with the %s formatter?

Short answer here is: Where you cannot use anything else. In C you don't have any support for complex datatypes such as a string. There are also no way of passing a variable "by reference" to a function. That's where you have to use pointers. Also you can have them to point at virtually anything, linked lists, members of structs and so on. But let's not go into that here.

The pointer operator (the *) is needed since we are telling printf that we want to print a character. Without the *, the character representation of the memory address itself would be printed. Now we are using the character itself instead. If we had used %s instead of %c, we would have asked printf to print the content of the memory address pointed to by 'a' plus one (in this example above), and we wouldn't have had to put the * in front:

But this would not have just printed the second character, but instead all characters in the next memory addresses, until a null character (\0) were found. And this is where things start to get dangerous. What if you accidentally try and print a variable of the type integer instead of a char pointer with the %s formatter?

With little effort and much confusion. ;-) If we talk about simple data types such as int and char there is little difference between an array and a pointer. These declarations are both the same

This would print whatever is found on memory address 120 and go on printing until a null character was found. It is not wrong or illegal to perform this printf statement, since a pointer actually is of the type int. Imagine the problems you might cause if you were to use sprintf() instead and assign this way too long "char array" to another variable, that only got a certain limited space allocated. You would most likely end up writing over something else in the memory and cause your program to crash (if you are lucky).

With little effort and much confusion. ;-) If we talk about simple data types such as int and char there is little difference between an array and a pointer. These declarations are very similar (but not the same - e.g., sizeof will return different values):

This would print whatever is found on memory address 120 and go on printing until a null character was found. It is wrong and illegal to perform this printf statement, but it would probably work anyway, since a pointer actually is of the type int in many environments. Imagine the problems you might cause if you were to use sprintf() instead and assign this way too long "char array" to another variable, that only got a certain limited space allocated. You would most likely end up writing over something else in the memory and cause your program to crash (if you are lucky).

char* a = "Hello"; int b = 120; printf("Second char is: %c", b); 

char* a = "Hello"; int b = 120; printf("Second char is: %s", b);